The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto. It is known (if I remember correctly) that the range isn't closed, but is dense in $C_0$. Everything I have read/heard says that the range is "difficult to describe".

But, since $\mathcal{F}$ is injective and continuous, the image of $L^1$ must be Borel inside of $C_0$. Is anything else known about its descriptive complexity? If not, might this be an example of a natural set of high Borel rank?

Isn't this space related to the Wiener Algebra? This is the space of $2\pi$-periodic functions with absolute summable Fourier coefficients. I vaguely remember that I heard somebody calling the image of $L^1$ under Fourier transform also "Wiener Algebra"...
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DirkApr 9 '13 at 8:46

Proof: Consider the cutoff functions $\{e^{-a\pi|t|}\}$ and fix a
sequence $a_n\to 0$. It is a fact that a function $g\in C_0(\mathbb
R)$ is the Fourier transform of some $f\in L^1$ if and only if the
sequence
\begin{equation}
T_n(g)(x) := \int_{\mathbb R} e^{-a_n\pi |t|} g(t)e^{2\pi itx} dt
\end{equation}
is Cauchy in $L^1$, in which case if we put $f=\lim T_ng$ then $g
=\widehat{f}$. Let $R$ denote the range of the Fourier transform in
$C_0$. If we define

Note that by monotone convergence, $g\in E_{m,n,k}$ if and only if two conditions are satisfied: first, for fixed $n$, we need $T_ng\in L^1$. By monotone convergence this is equivalent to: There exists an integer $N$ such that for all integers $d\geq 1$,

By dominated convergence the set of all such $g$ obeying this for fixed $N,n,d$ is closed, so the set that obeys this for some $N$ and all $d$ (with $n$ held fixed) is an $F_\sigma$.
Thus for fixed $m,n$ the set of $g$ for which $T_ng, T_mg\in L^1$ is an $F_\sigma$.
Additionally, for all integers $d\geq 1$, we need the condition

It seems that Mike's argument shows that in fact the range $R$ of the Fourier transform is $F_{\sigma\delta}$ in $C_0(\mathbb R)$. Define the $T_ng$ as above,
$$T_ng(x)=\int_{\mathbb R} e^{-a_n\pi\vert t\vert}g(t) e^{2i\pi tx}dt\;.$$ Then a function $g\in C_0(\mathbb R)$ is in $R$ iff two things hold:

Altogether, $R$ is the intersection of two $F_{\sigma\delta}$ sets, hence an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$. I would be extremely surprised if it were better than that; i.e. I "conjecture" that it is not $G_{\delta\sigma}$.